Powering-invariance is not quotient-transitive

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This article gives the statement, and possibly proof, of a subgroup property (i.e., powering-invariant subgroup) not satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup property).
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Statement

It is possible to have groups H \le K \le G such that H is a powering-invariant normal subgroup of G and K/H is a powering-invariant subgroup of the quotient group G/H, but K is not powering-invariant in G.

Related facts

Proof

The proof idea is follows: use the construction in the reference for H. Now take K as a subgroup containing H such that K/H is a finite cyclic group of order n > 1. Now:

  • H is powering-invariant in G by construction, since both G and H are rationally powered groups.
  • K/H is powering-invariant in G/H since G/H is not powered over any prime.
  • K is not powering-invariant in G: For instance, an element of K whose image in K/H generates the latter group does not have a n^{th} root in K.

References